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lyudmila [28]
2 years ago
13

An accelerometer is a device that uses the extension of a spring to measure acceleration in terms of Earth's gravitational accel

eration (g). What is the approximate acceleration if this accelerometer spring is extended just 0.43 centimeters?. . a) 0.24 g. b) 0.80 g. c) 2.00 g. d) 2.40 g
Physics
2 answers:
Nimfa-mama [501]2 years ago
8 0

0.80g (B) is correct for plato!

ELEN [110]2 years ago
3 0
In your question explains that an accelerometer is a device that uses the extension spring to measure acceleration in terms of Earth's Gravitational accelerations. So if this device is extended just 0.43 centimeters the possible accelerations is letter B. 0.80 g
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If a train is 100 kilometers away, how much sooner would you hear the train coming by listening to the rails (iron) as opposed t
Whitepunk [10]
From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)

Distance traveled is
d = 100 km = 10⁵ m

Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s

Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s

The difference in time is
302.12 - 19.49 = 282.63 s

Answer:  283 s (nearest second)



6 0
2 years ago
Two boxes are connected to each other by a string as shown in the figure. The 10-n box slides without friction on the horizontal
FromTheMoon [43]

Answer:

T=7.4 N hence T<30 N

Explanation:

The figure is likely to be similar to the one attached. Writing the equation for forces we have

F-T=Fa/g where F is the force, T is tension, a is acceleration and g is acceleration due to gravity. Substituting the figures we have the first equation as

30 N - T = (30/9.81)a

Also, we know that T=F*a/g and substituting  10N for F we obtain the second equation as

T = (10/9.81)a

Adding the first and second equations we obtain

30 = 4.077471967

a Hence

a=\frac {30}{4.077471967}=7.3575 m/s^{2}

and T=a hence

T is approximately 7.4 N

5 0
2 years ago
Read 2 more answers
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

4 0
2 years ago
At room temperature, a typical person loses energy to the surroundings at the rate of 62 W. If this energy loss has to be made u
Alex_Xolod [135]

To solve this problem it is necessary to use the given proportions of power and energy, as well as the energy conversion factor in Jules to Calories.

The power is defined as the amount of energy lost per second and whose unit is Watt. Therefore the energy loss rate given in seconds was

P = \frac{E}{t} \rightarrow E= Energy, t = time

P = 62W = 62 \frac{J}{s}

The rate of energy loss per day would then be,

P = 62\frac{J}{s} (\frac{86400s}{1day})

P = 5356800 \frac{J}{day}

That is to say that Energy in Jules per lost day is 5356800J

By definition we know that 1KCal = 4.184*10^{6}J

In this way the energy in Cal is,

E = 5356800J \frac{1KCal}{4.184*10^{6}J}

E = 1279.694 KCal

The number of kilocalories (food calories) must be 1279.694 KCal

4 0
2 years ago
The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes
Elodia [21]

Explanation:

Below is an attachment containing the solution.

7 0
2 years ago
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